Problem: Which of the definite integrals is equivalent to the following limit? $ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4+\dfrac{5i}n}\cdot\dfrac5n$ Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^5 \sqrt {x}\,dx$ (Choice B) B $ \int_4^9 \sqrt {4+x}\,dx$ (Choice C) C $ \int_4^9 \sqrt {x}\,dx$ (Choice D) D $ \int_0^4 \sqrt {4+x}\,dx$
Solution: The value of a definite integral is the limit of its Riemann sums as the number of terms tends to infinity. The given summation $ \sum_{i=1}^n \sqrt{4+\dfrac{5i}n}\cdot\dfrac5n$ looks like a right Riemann sum with $n$ subintervals of equal width. If each subinterval has width $\Delta x$, what is the right Riemann sum for the following definite integral? $ \int_a^b f(x) \,dx$ The right Riemann sum for the definite integral is $ \sum_{i=1}^n f(a+i\Delta x)\cdot\Delta x$. What does the sum become when we express $\Delta x$ in terms of $a$, $b$, and $n$ ? Dividing the interval $[a,b]$ into $n$ subintervals of equal width yields a common width of $\Delta x=\dfrac{b-a}n\,$. This lets us express the right Riemann sum as $ \sum_{i=1}^n f\left(a+i\cdot\dfrac{b-a}n\right)\cdot\dfrac{b-a}n$. Let's rewrite the given summation as $ \sum_{i=1}^n \sqrt{4+i\cdot\dfrac5n}\cdot\dfrac5n\,$. If the given summation is a right Riemann sum, what are $a$, $b$, and $f$ ? Equating the width $\Delta x$ of each subinterval in the two sums yields $\dfrac{b-a}n=\dfrac5n\,$. Thus, the interval $[a,b]$ has width $b-a=5$. In two of the answer choices, $f(x)=\sqrt x$. In that case, $a=4$ and $b=a+5=9$. These values produce the definite integral $ \int_4^9 \sqrt x \,dx$. In two other answer choices, $f(x)=\sqrt{4+x}$. In that case, $a=0$ and $b=a+5=5$. These values produce the definite integral $ \int_0^5 \sqrt{4+x} \,dx$. Are either of these two integrals answer choices? The correct answer is $ \lim_{n\to\infty} \sum_{i=1}^n \sqrt{4+\dfrac{5i}n}\cdot\dfrac5n=\int_4^9 \sqrt x \,dx$.